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Chapter 5: Problem 37

Verify each identity. $$\frac{\sin ^{2} x-\cos ^{2} x}{\sin x+\cos x}=\sin x-\cos x$$

Short Answer

Expert verified
After simplifying the left-hand side of the identity to \( \sin x - \cos x \), both sides of the equation are equal, proving the identity

Step by step solution

01

Rewrite numerator using difference of squares

The numerator \( \sin^{2}x - \cos^{2}x \) on the left-hand side of the given identity can be written as a difference of squares: \( (\sin x + \cos x)(\sin x - \cos x) \)
02

Simplify the left-hand side

Since the denominator of the left-hand side of the identity is \( \sin x + \cos x \), we can divide it by the same factor from the numerator, which leaves us with \( \sin x - \cos x \)
03

Compare

Now both sides of the identity are the same: \( \sin x - \cos x = \sin x - \cos x \). This confirms that the given expression is indeed an identity

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Most popular questions from this chapter

Verify the identity: $$\frac{\sin (x-y)}{\cos x \cos y}+\frac{\sin (y-z)}{\cos y \cos z}+\frac{\sin (z-x)}{\cos z \cos x}=0$$

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Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$5 \cot ^{2} x-15=0$$
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